两个数字的倍数排序列表中的第N个倍数

给定三个正整数a，b和n 。考虑一个具有“ a”和“ b”的所有倍数的列表。该列表已排序，并删除了重复项。任务是找到列表的第n个元素。

例子 ：

Input :  a = 3, b = 5, n = 5
Output : 10
a = 3 b = 5, Multiple of 3 are 3, 6, 9, 12, 15,... and 
multiples of 5 are 5, 10, 15, 20,.... 
After deleting duplicate element and sorting:
3, 5, 6, 9, 10, 12, 15, 18, 20,.... 
The 5th element in the sequence is 10.

Input : n = 6, a = 2, b = 3 
Output : 9

方法1 ：(强力)
生成’a’的第一个’n’倍数。现在，生成b的第一个’n’倍数，使其不属于’a’的前n个倍数。可以使用二进制搜索来完成。

C++

// C++ program to find n-th number in the sorted
// list of multiples of two numbers.
#include
using namespace std;
  
// Return the n-th number in the sorted
// list of multiples of two numbers.
int nthElement(int a, int b, int n)
{
    vector seq;
  
    // Generating first n multiple of a.
    for (int i = 1; i <= n; i++)
        seq.push_back(a*i);
  
    // Sorting the sequence.
    sort(seq.begin(), seq.end());
  
    // Generating and storing first n multiple of b
    // and storing if not present in the sequence.
    for (int i = 1, k = n; i <= n && k; i++)
    {
        // If not present in the sequence
        if (!binary_search(seq.begin(), seq.end(), b*i))
        {
            // Storing in the sequence.
            seq.push_back(b*i);
  
            sort(seq.begin(), seq.end());
            k--;
        }
    }
  
    return seq[n - 1];
}
  
// Driven Program
int main()
{
    int a = 3, b = 5, n = 5;
    cout << nthElement(a, b, n) << endl;
    return 0;
}

Java

// Java program to find n-th number 
// in the sorted list of multiples
// of two numbers.
import java.io.*; 
import java.util.*; 
class GFG
{
// Return the n-th number in the sorted
// list of multiples of two numbers.
static int nthElement(int a, int b, int n)
{
    ArrayList seq = new
              ArrayList(n * n + 1); 
      
    // Generating first n multiple of a.
    for (int i = 1; i <= n; i++)
        seq.add(a * i);
          
    // Sorting the sequence.
    Collections.sort(seq);
      
    // Generating and storing first n 
    // multiple of b and storing if 
    // not present in the sequence.
    for (int i = 1, k = n; 
             i <= n && k > 0; i++)
    {
        // If not present in the sequence
        if (seq.indexOf(b * i) == -1)
        {
            // Storing in the sequence.
            seq.add(b * i);
            Collections.sort(seq);
            k--;
        }
    }
    return seq.get(n - 1);
}
  
// Driver Code
public static void main(String[] args)
{
    int a = 3, b = 5, n = 5;
    System.out.println(nthElement(a, b, n));
}
}
  
// This code is contributed by mits

Python3

# Python3 program to find n-th 
# number in the sorted list of 
# multiples of two numbers.
  
# Return the n-th number in the
# sorted list of multiples of 
# two numbers.
def nthElement(a, b, n):
    seq = [];
  
    # Generating first n 
    # multiple of a.
    for i in range(1, n + 1):
        seq.append(a * i);
  
    # Sorting the sequence.
    seq.sort();
  
    # Generating and storing first 
    # n multiple of b and storing
    # if not present in the sequence.
    i = 1;
    k = n; 
    while(i <= n and k > 0):
          
        # If not present in the sequence
        try:
            z = seq.index(b * i);
        except ValueError:
              
            # Storing in the sequence.
            seq.append(b * i);
            seq.sort();
            k -= 1;
        i += 1;
  
    return seq[n - 1];
  
# Driver Code
a = 3;
b = 5;
n = 5;
print(nthElement(a, b, n));
  
# This code is contributed by mits

C#

// C# program to find n-th number
// in the sorted list of multiples
// of two numbers.
using System;
using System.Collections;
  
class GFG
{
// Return the n-th number in the sorted
// list of multiples of two numbers.
static int nthElement(int a,
                      int b, int n)
{
    ArrayList seq = new ArrayList(); 
      
    // Generating first n multiple of a.
    for (int i = 1; i <= n; i++)
        seq.Add(a * i);
  
    // Sorting the sequence.
    seq.Sort();
  
    // Generating and storing first n 
    // multiple of b and storing if
    // not present in the sequence.
    for (int i = 1, k = n; 
             i <= n && k > 0; i++)
    {
        // If not present in the sequence
        if (!seq.Contains(b * i))
        {
            // Storing in the sequence.
            seq.Add(b * i);
            seq.Sort();
            k--;
        }
    }
  
    return (int)seq[n - 1];
}
  
// Driver Code
static void Main()
{
    int a = 3, b = 5, n = 5;
    Console.WriteLine(nthElement(a, b, n));
}
}
  
// This code is contributed by mits

PHP

 0; $i++)
    {
        // If not present in the sequence
        if (array_search($b * $i, $seq) == 0)
        {
            // Storing in the sequence.
            array_push($seq, $b * $i);
   
            sort($seq);
            $k--;
        }
    }
  
    return $seq[$n - 1];
}
  
// Driver Code
$a = 3;
$b = 5;
$n = 5;
echo nthElement($a, $b, $n);
  
// This code is contributed by mits
?>

输出：

10

方法2 ：(有效方法)
想法是利用a和b的公倍数通过LCM(a，b)删除这一事实。令f(a，b，x)是一个函数，该函数计算小于x以及a和b的倍数的数量。现在，使用包含-排除原理，我们可以说：

f(a, b, x) :  Count of number that are less than x
              and multiples of a and b

f(a, b, x) = (x/a) + (x/b) - (x/lcm(a, b))
where (x/a) define number of multiples of a
(x/b) define number of multiple of b
(x/lcm(a, b)) define the number of common multiples 
of a and b.

观察到，a和b是恒定的。随着x的增加，f(a，b，x)也将增加。因此，我们可以应用二元搜索来找到x的最小值，使得f(a，b，x)> = n。函数的下限是必需的答案。

第n个项的上限为min(a，b)* n。请注意，当a和b的倍数中没有公共元素时，我们得到第n个值的最大值。

以下是上述方法的实现：

C++

// C++ program to find n-th number in thes
// sorted list of multiples of two numbers.
#include
using namespace std;
  
// Return the Nth number in the sorted
// list of multiples of two numbers.
int nthElement(int a, int b, int n)
{
    // Finding LCM of a and b.
    int lcm = (a * b)/__gcd(a,b);
  
    // Binary Search.
    int l = 1, r = min(a, b)*n;
    while (l <= r)
    {
        // Finding the middle element.
        int mid = (l + r)>>1;
  
        // count of number that are less than
        // mid and multiples of a and b
        int val = mid/a + mid/b - mid/lcm;
  
        if (val == n)
            return max((mid/a)*a, (mid/b)*b);
  
        if (val < n)
            l = mid + 1;
        else
            r = mid - 1;
    }
}
  
// Driven Program
int main()
{
    int a = 5, b = 3, n = 5;
    cout << nthElement(a, b, n) << endl;
    return 0;
}

Java

// Java program to find n-th number in the
// sorted list of multiples of two numbers.
import java.io.*;
  
public class GFG{
      
// Recursive function to return 
// gcd of a and b
static int __gcd(int a, int b)
{
    // Everything divides 0 
    if (a == 0 || b == 0)
    return 0;
  
    // base case
    if (a == b)
        return a;
      
    // a is greater
    if (a > b)
        return __gcd(a - b, b);
        return __gcd(a, b - a);
}
  
// Return the Nth number in the sorted
// list of multiples of two numbers.
static int nthElement(int a, int b, int n)
{
    // Finding LCM of a and b.
    int lcm = (a * b) / __gcd(a, b);
  
    // Binary Search.
    int l = 1, r = Math.min(a, b) * n;
    while (l <= r)
    {
        // Finding the middle element.
        int mid = (l + r) >> 1;
  
        // count of number that are less than
        // mid and multiples of a and b
        int val = mid / a + mid / b - 
                  mid / lcm;
  
        if (val == n)
            return Math.max((mid / a) * a, 
                            (mid / b) * b);
  
        if (val < n)
            l = mid + 1;
        else
            r = mid - 1;
    }
    return 0;
}
  
// Driver Code
static public void main (String[] args)
{
    int a = 5, b = 3, n = 5;
    System.out.println(nthElement(a, b, n));
}
}
  
// This code is contributed by vt_m. 

的Python 3

# Python 3 program to find n-th number
# in thes sorted list of multiples of 
# two numbers.
import math
  
# Return the Nth number in the sorted
# list of multiples of two numbers.
def nthElement(a, b, n):
  
    # Finding LCM of a and b.
    lcm = (a * b) / int(math.gcd(a, b))
  
    # Binary Search.
    l = 1
    r = min(a, b) * n
    while (l <= r):
      
        # Finding the middle element.
        mid = (l + r) >> 1
  
        # count of number that are less
        # than mid and multiples of a 
        # and b
        val = (int(mid / a) + int(mid / b)
                         - int(mid / lcm))
  
        if (val == n):
            return int( max(int(mid / a) * a,
                          int(mid / b) * b) )
  
        if (val < n):
            l = mid + 1
        else:
            r = mid - 1
      
# Driven Program
a = 5
b = 3
n = 5
print(nthElement(a, b, n))
  
# This code is contributed by Smitha.

C＃

// C# program to find n-th number in thes
// sorted list of multiples of two numbers.
using System;
  
public class GFG{
      
// Recursive function to return
// gcd of a and b
static int __gcd(int a, int b)
{
    // Everything divides 0 
    if (a == 0 || b == 0)
    return 0;
  
    // base case
    if (a == b)
        return a;
      
    // a is greater
    if (a > b)
        return __gcd(a - b, b);
        return __gcd(a, b - a);
}
  
// Return the Nth number in the sorted
// list of multiples of two numbers.
static int nthElement(int a, int b, int n)
{
    // Finding LCM of a and b.
    int lcm = (a * b) / __gcd(a, b);
  
    // Binary Search.
    int l = 1, r = Math.Min(a, b) * n;
    while (l <= r)
    {
        // Finding the middle element.
        int mid = (l + r) >> 1;
  
        // count of number that are less than
        // mid and multiples of a and b
        int val = mid / a + mid / b - 
                  mid / lcm;
  
        if (val == n)
            return Math.Max((mid / a) * a, 
                            (mid / b) * b);
  
        if (val < n)
            l = mid + 1;
        else
            r = mid - 1;
    }
    return 0;
}
  
// Driver Code
    static public void Main (String []args)
    {
        int a = 5, b = 3, n = 5;
            Console.WriteLine(nthElement(a, b, n));
    }
}
  
// This code is contributed by vt_m. 

的PHP

> 1;
  
        // count of number that are 
        // less than mid and multiples
        // of a and b
        $val = (int)($mid / $a) + 
               (int)($mid / $b) - 
               (int)($mid / $lcm);
  
        if ($val == $n)
            return max((int)(($mid / $a)) * $a, 
                       (int)(($mid / $b)) * $b);
  
        if ($val < $n)
            $l = $mid + 1;
        else
            $r = $mid - 1;
    }
}
  
// Driver code
$a = 5;
$b = 3;
$n = 5;
echo nthElement($a, $b, $n);
  
// This code is contributed by mits 
?>

输出：

10